What is the time complexity of Minimum Insertion Steps to Make a String Palindrome?

The optimal solution for Minimum Insertion Steps to Make a String Palindrome runs in O(n²) time and uses O(n²) space. The time complexity is O(n²) due to the nested loops filling the DP table. The space complexity is also O(n²) because we store results in a 2D array.

What is the brute force solution for Minimum Insertion Steps to Make a String Palindrome?

The brute force approach for Minimum Insertion Steps to Make a String Palindrome is Brute Force, with O(2^n) time complexity and O(n) space complexity. The brute-force approach involves checking all possible ways to insert characters to make the string a palindrome. This can be done by recursively trying to insert characters at different positions and counting the steps needed. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Minimum Insertion Steps to Make a String Palindrome?

Minimum Insertion Steps to Make a String Palindrome uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Minimum Insertion Steps to Make a String Palindrome?

Explain your thought process clearly while coding, especially when transitioning from brute force to optimal solutions. Practice writing out the DP table and understanding how it evolves with each step. Be prepared to discuss the trade-offs between different approaches and why you chose one over the other.

What are common mistakes when solving Minimum Insertion Steps to Make a String Palindrome?

Not considering base cases in recursive solutions, leading to incorrect results. Failing to optimize the recursive solution, resulting in excessive time complexity.

#1312

Minimum Insertion Steps to Make a String Palindrome

Hard

StringDynamic ProgrammingDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^n)	O(n²)
Space	O(n)	O(n²)

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Intuition

Time O(n²)Space O(n²)

The optimal approach uses dynamic programming to find the longest palindromic subsequence (LPS) in the string. The minimum insertions required will be the difference between the string length and the LPS length.

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Algorithm

3 steps

1Step 1: Initialize a 2D DP array where dp[i][j] represents the length of the longest palindromic subsequence in the substring s[i:j].
2Step 2: Fill the DP table: if s[i] == s[j], then dp[i][j] = dp[i + 1][j - 1] + 2; otherwise, dp[i][j] = max(dp[i + 1][j], dp[i][j - 1]).
3Step 3: The answer is the length of the string minus the value in dp[0][n-1], where n is the length of the string.

solution.py13 lines

1def min_insertions(s):
2    n = len(s)
3    dp = [[0] * n for _ in range(n)]
4    for i in range(n):
5        dp[i][i] = 1
6    for length in range(2, n + 1):
7        for i in range(n - length + 1):
8            j = i + length - 1
9            if s[i] == s[j]:
10                dp[i][j] = dp[i + 1][j - 1] + 2
11            else:
12                dp[i][j] = max(dp[i + 1][j], dp[i][j - 1])
13    return n - dp[0][n - 1]

ℹ

Complexity note: The time complexity is O(n²) due to the nested loops filling the DP table. The space complexity is also O(n²) because we store results in a 2D array.

1Understanding the relationship between the longest palindromic subsequence and the minimum insertions needed.
2Dynamic programming can significantly reduce the complexity of problems involving subsequences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.